V(t) = 100 + (5t+4) - t^2/2
so you are correct there.
V'(t) = 5-t
So, you are correct that V is a max at t=5
The average value of a function is
(∫[a,b] f(t) dt)/(b-a) So, that means that #3 is
∫[0,5] (-0.5t^2+5t+104) dt = 337/3
So, you want to find when
(-t^2/2 + 5t + 104) = 337/3
t = 5 - 5/√3 ≈ 2.113
I've been trying to solve this problem for the past 45 minutes and could really use some help. I'm given that water flows into a tank at a rate of (5t+4) gallons/min. Water flows out of the tank at a rate of 0.5(t^2) gallons/min. At t=0 min the tank contains 100 gallons.
The questions are: 1) Write an expression for the amount of water in the tank at any given time t
2) When will the quantity of water in the tank be a maximum and how much is the quantity of water at the time? Show the work that leads to your answers.
3) Find the average number of gallons of water in the tank in the first five minutes, and determine the time during the first 5 minutes when this average number is actually obtained. Justify your answers.
What I have so far (and I'm pretty sure it's worng) is 1) V(t)= -0.5t^2+5t+104
2) V'(t)=-t + 5 solve for V'(0) and got 5.
3) Solved for V(5) and plugged that answer back into V'(t). So, I got V(5)= 116.5 and then plugged it in to get V'(116.5)= -111.5
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